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Standard VIII

Physics

Buoyant Force

A piece of wa...

Question

A piece of wax weights $18.03 g$ in air. A piece of metal is found to weigh $17.03 g$ in water. It is tied to the wax and both together weigh $15.23 g$ in water. Then, the specific gravity of wax is

\frac{18.03}{17.03}

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\frac{17.03}{18.03}

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\frac{18.03}{19.83}

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\frac{15.03}{17.03}

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Solution

The correct option is C $\frac{18.03}{19.83}$
$W_{(w a x)} = 18.03 g$
In water,
$15.23 g = 17.03 + (18.03 - V \frac{ρ_{(w a x)}}{ρ_{w a x}} \times ρ_{w} g) 15.23 = 17.03 + (18.03 - \frac{18.03}{s p e c i f i c g r a v i t y}) s p e c i f i c g r a v i t y = \frac{18.03}{19.83}$

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A piece of wax weights 18.03g in air. A piece of metal is found to weigh 17.03g in water. It is tied to the wax and both together weigh 15.23g in water. Then, the specific gravity of wax is

The correct option is C 18.0319.83 W(wax)=18.03gIn water, 15.23g=17.03+(18.03−Vρ(wax)ρwax×ρwg)15.23=17.03+(18.03−18.03specificgravity)specificgravity=18.0319.83

A piece of wax weights $18.03 g$ in air. A piece of metal is found to weigh $17.03 g$ in water. It is tied to the wax and both together weigh $15.23 g$ in water. Then, the specific gravity of wax is

The correct option is C $\frac{18.03}{19.83}$
$W_{(w a x)} = 18.03 g$
In water,
$15.23 g = 17.03 + (18.03 - V \frac{ρ_{(w a x)}}{ρ_{w a x}} \times ρ_{w} g) 15.23 = 17.03 + (18.03 - \frac{18.03}{s p e c i f i c g r a v i t y}) s p e c i f i c g r a v i t y = \frac{18.03}{19.83}$